I assume your display part is OK, and you already verified, that the interrupt routine correctly displays count1 as the rightmost digit, count2, count3 and then count4 as the leftmost digit (although I would NOT display all 4 digits within the interrupt routine and use delays; rather, display only 1 digit at a time, and advance to the next digit the next time the timer interrupt kicks in).

I think the main problem is at the end of the scan routine:

done:
setb c
mov a,r6
mov unit,a
mov count1,a
xch a,count1
mov count1,a
xch a,count2
mov count2,a
mov a,count2
xch a,count3
mov count3,a
mov a,count3
xch a,count4
mov count4,a
exit:
ret 

First, this should NOT be at the end of the keys scan routine - if the key is pressed, it is called many many times and so the display is almost immediately filled (all 4 positions) with the currently pressed key. You should move the "add one character to the display" (shift 1 digit left) routine to after the debounding, before the release test (just before "back2" label - and you don't need the push acc/pop acc there).

And I think this is completely defective - have you tried it in a simulator? I think it should go as follows (many people don't know that '51 is NOT a typical accumulator based architecture and some operations can be done directly on the internal memory positions, move is one of them - any two directly addressed positions can be moved from and to):

; assuming valid key code in accumulator
  mov  count4,count3
  mov  count3,count2
  mov  count2,count1
  mov  count1,a

PS. Next time, please enclose the code sample between <pre> and </pre> tags to preserve formatting.